數學題permutation and combination

invisibleman

2010-12-31 2:11 am

1 ) 6 boys and 4 girls are lined up a) If girls are at both ends , how many permutation are there ?ans : 4P2 x 8 ! = 4838404P2 < ( arrange 2 girls at both end ) 8! < (Arrange the remaining students )問題係 If girls are at both ends 唔係話兩邊都有兩個女仔咩 ?因為有4個女仔 ... 4P2 只係arrange 2 girls at both end 所以唔係應該 4P2 x 2 x 8! ?__________________________________________________________________________________________________2)There are six different candies.a )How many ways are there to split the candies into three piles , such that there is one candy in one pile ,two candies in another , and three candies ANS : 6C1 x 5C2 x 3C3 呢個我明 .. 但...b) how many ways are there o distribute the candies to Albert, Benny and Carl , such that one of them can get one candy ,another one gets two , and the remaining one gets three ?ANS :6C1 x 5C2 x 3C3 x 3! 就係唔明點解要乘返3 ! 咁 distribute to piles 同 3 個唔同既人.. 都係咁 distribute , 咁點解仲要乘3! ???有咩分別 ?c) how many ways are there to split the candies into three piles , such that there are two candies in each pile ANS : (6C2 x 4C2 x 2C2) / 3! 點解要除返 3! 呢 ?唔該晒 !

更新1:

thx a lot ! one more question : There are 10 seats in a row in a waiting room . There are six people in the room. a) In the group of six people , there are 3 sisters who must sit next to each other .In how many different ways can the group be seated ? ANS : 10080 how to do this ?

回答 (1)

myisland8132

2010-12-31 2:32 am

✔ 最佳答案

就版大的三個問題回答如下：1 因為P已經有排列的功能。例如若果該兩個位置是叫1﹐2。又抽到的GIRL是A,B﹐則1A2B和1B2A在4P2中已經分開處理了﹐所以不用再乘2。2 因為該三堆東西給人的感覺是完全不同的。例如M1-1個candy‧M2-2個candy。M3-3個candy。則 Albert-M1, Benny-M2 and Car-M3 和 Albert-M3, Benny-M2 and Car-M1 好明顯不同。頭一個Car高興Albert不高興。後一個則相反。所以你要計入這三個人的排列3!=6。3 舉個例。假如你先拿AB兩個candy塞入G1﹐然後CD塞入G2﹐最後EF塞入G3。這是一種分配法。第二次你先拿EF兩個candy塞入G1﹐然後CD塞入G2﹐最後AB塞入G3﹐這是另一種分配法。但若果G1,G2,G3是同一種牌子的膠袋的話﹐則2種分配法是沒有分別的。所以真正的分配數要除回G1,G2,G3的排列數3!=6。

2010-12-31 14:29:05 補充：
4 For 10 seats , For 10 seats , there are 8 groups of 3 seats next each other.

3 sisters sit next to each other in 10 seats : (3P3) * 8 = 48 ways ,

remain 7 seats for other 3 people have 7P3 = 210 ways ,

The number of different ways can the group be seated = 48 * 210 = 10080

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